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I'm trying to prove the following: $$M_{X}(t)=\prod\limits_{i=1}^n M_{X_i}\left(\frac{t}{n}\right)$$ Where $X := \frac{1}{n} \sum\limits_{i=1}^n X_i $ and all the random variables are independent.

All I need to finish the prove is to show that $e^{\frac{t}{n}X_i}$ are independent which I'm not sure how to do. I was thinking in general if it is true that the bijection of independent random variables are also independent i.e. if we biject $X_i$ to $Y_i$, are the $Y_i$ independent given that the $X_i$ are?

Regards, Raxel.

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  • $\begingroup$ It depends on what definition of independence you have seen. In general, more is true. If $X$ and $Y$ are independent random variables, then $g(X)$ and $h(Y)$ are also independent for any (measurable) functions $h$ and $g$. $\endgroup$
    – passerby51
    Commented Aug 29, 2013 at 22:02
  • $\begingroup$ Ok is there any proof that you know of on the web? $\endgroup$
    – Raxel
    Commented Aug 29, 2013 at 22:04
  • $\begingroup$ I am not sure. But if you know a particular definition of independence, it is almost immediate. Let me write it as an answer. $\endgroup$
    – passerby51
    Commented Aug 29, 2013 at 22:07
  • $\begingroup$ Somebody should probably mention that the phrases "the bijection of independent random variables" and "to biject [some random variable] to [some other random variable]" do not refer to any known notion. $\endgroup$
    – Did
    Commented Aug 31, 2013 at 7:36

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One can define two random variables $X$ and $Y$ to be independent if (and only if) the following holds $$ E [u(X) v(Y)] = E[u(X)]\; E[v(Y)] \quad (*) $$ for any (measurable) function $u$ and $v$. Note that a special case of this is when $u(x) = 1\{ x \in A\}$ and $v(y) = \{y \in B\}$ for (meas.) sets $A$ and $B$. That is, if $u$ and $v$ are indicator functions. Then, the above reduces to $$P( X \in A, Y \in B) = P(X \in A) P(Y \in B)$$ which might be the definition that you have seen. (Though it may seem that this is a special case, it is in fact equivalent to the above mentioned ($*$))

Assuming ($*$), it is easy to see that if $X$ and $Y$ are independent, then $g(X)$ and $h(Y)$ are also independent (that is, they too satisfy ($*$)), since the composition of measurable functions is again a measurable function.

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    $\begingroup$ One could also note that $[g(X)\in A,h(Y)\in B]=[X\in C,Y\in D]$ with $C=g^{-1}(A)$ and $D=h^{-1}(B)$, and proceed from here. But I prefer the $(\ast)$ route (and I think this "functional" approach should be as much advertised as possible). $\endgroup$
    – Did
    Commented Aug 30, 2013 at 6:45

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